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Worked solutions

IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 16

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–16 · 110 marks

Question 1

5 marks

Exponential population model

Functions

(a)

1 mark

At the start of 2024, $t = 0$ and $b^{0} = 1$ , so $a$ is just the starting population.

f (0) = a \times b^{0} = a a = 104000

Mentor tip

The value of $a$ in $a \times b^{t}$ is always the value at $t = 0$ .

(b)

2 marks

One year later $t = 1$ . Set $f (1)$ equal to the 2025 figure and divide.

104000 \times b = 105144 b = \frac{105 144}{104 000} = 1.011

Mentor tip

The word exact matters here: give $1.011$ in full, not a rounded value.

(c)

2 marks

The start of 2034 is ten years on, so use $t = 10$ .
Round to the nearest whole person, as asked.

f (10) = 104000 \times 1.01 1^{10} = 116 023.2 \dots \approx 116023

Mentor tip

Count the years carefully: 2024 is $t = 0$ , so 2034 is $t = 10$ , not $t = 11$ .

Question 2

5 marks

Midpoint and distance in 3D

Geometry and trigonometry

(a)

2 marks

Average each coordinate of $A$ and $B$ .

F = (\frac{0 + 40}{2}, \frac{60 + 0}{2}, \frac{240 + 240}{2}) F = (20, 30, 240)

Mentor tip

Write the answer as a bracketed coordinate triple, not three loose numbers.

(b)

3 marks

Use the 3D distance formula between $S (40, 0, 160)$ and $F (20, 30, 240)$ .

S F = (40 - 20)^{2} + (0 - 30)^{2} + (160 - 240)^{2} = 400 + 900 + 6400 = 7700 \approx 87.7 cm

Another way. Find $F B = 2 0^{2} + 3 0^{2}$ across the level of the ceiling thread, then combine it with the vertical drop $S B = 80$ using Pythagoras.

Mentor tip

The distance formula is in the formula booklet. The squares remove any sign worries.

Question 3

9 marks

Two-way table and a chi-squared test

Statistics and probability

(a)

1 mark

Read the single cell for under 30 and cold, out of all $180$ people.

\frac{62}{180} = \frac{31}{90} \approx 0.344

(b)

2 marks

The condition shrinks the group to the over-50 column only: $19 + 34 = 53$ people.
Of those, $19$ prefer hot.

\frac{19}{53} \approx 0.358

Mentor tip

For a conditional probability the denominator is the column total, not $180$ .

(c)

1 mark

The null hypothesis for a test of independence always says the two variables are independent.

H_{0} : age and preferred type of drink are independent

Mentor tip

Use the word independent. Saying the variables are not related or not correlated is too loose.

(d)

3 marks

Enter the $2 \times 3$ table of observed counts as a matrix and run the chi-squared test on the GDC.
Read off both the statistic and the $p$ -value. There are $(2 - 1) (3 - 1) = 2$ degrees of freedom.

(i) χ_{calc}^{2} = 3.65997 \dots \approx 3.66 (GDC) (ii) p = 0.160415 \dots \approx 0.160 (GDC)

Mentor tip

Enter only the six observed counts, not the row or column totals.

(e)

2 marks

Compare the statistic with the critical value (or the $p$ -value with $0.01$ ).
The statistic is smaller, so there is not enough evidence to reject $H_{0}$ .

3.66 < 9.210 (or 0.160 > 0.01) Do not reject H_{0} : age and drink preference are independent.

Mentor tip

Always write the comparison first, then the conclusion. A conclusion with no reason is only half an answer.

Question 4

4 marks

Linear model from two bills

Number and algebra

4 marks

The bill is linear: bill $= c + u \times$ (units used).
Write one equation for each month and solve them simultaneously (GDC or subtraction).
(i) and (ii): subtracting the equations removes $c$ and gives $u$ ; substitute back for $c$ .

c + 3000 u = 117.90 c + 4800 u = 179.10 1800 u = 61.20 ⟹ u = 0.034 c = 117.90 - 3000 (0.034) = 15.90 (i) c = $15.90 (ii) u = $0.034

Mentor tip

Check both bills with your values: $15.90 + 4800 \times 0.034 = 179.10$ .

Question 5

7 marks

Arithmetic sequence: a term and a sum

Number and algebra

(a)

2 marks

Use $u_{n} = u_{1} + (n - 1) d$ with $u_{1} = 18$ , $d = 7$ and $n = 42$ .

u_{42} = 18 + 41 \times 7 = 305

Mentor tip

It is $41$ steps of $7$ to reach the 42nd term, not $42$ .

(b)

3 marks

Put the values into $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ and set it equal to $9475$ .
Expand and move everything to one side.

\frac{n}{2} (36 + 7 (n - 1)) = 9475 \frac{n}{2} (7 n + 29) = 9475 7 n^{2} + 29 n - 18950 = 0

Mentor tip

Halving every term, $3.5 n^{2} + 14.5 n - 9475 = 0$ , is just as good.

(c)

2 marks

Solve the quadratic with the GDC polynomial solver.
Keep only the positive whole-number root, since $n$ counts terms.

n = 50 or n = - 54.14 \dots (GDC) n = 50

Mentor tip

Reject the negative root in writing and give one final answer.

Question 6

7 marks

Closest point on a line to a point

Geometry and trigonometry

7 marks

The closest point is where a perpendicular from $C$ meets the road.
Find the gradient and equation of the road through $A$ and $B$ .
The perpendicular through $C$ has the negative reciprocal gradient.
Solve the two line equations simultaneously to get $D$ .

m_{A B} = \frac{2 - 10}{5 - 1} = - 2 road: y = - 2 x + 12 m_{C D} = \frac{1}{2} : y - 5 = \frac{1}{2} (x - 4) ⟹ y = 0.5 x + 3 - 2 x + 12 = 0.5 x + 3 ⟹ x = 3.6 D = (3.6, 4.8)

Another way. Write the distance from $C$ to a general point $(x, - 2 x + 12)$ on the road and use the GDC to find the $x$ that minimises it.

Mentor tip

Check $D$ lies on the road: $- 2 (3.6) + 12 = 4.8$ .

Question 7

4 marks

Euler's method for a differential equation

Calculus

4 marks

Use $y_{n + 1} = y_{n} + 0.1 (x_{n} + ln (1 + x_{n}))$ and $x_{n + 1} = x_{n} + 0.1$ , starting at $(1, 4)$ .
Five steps take $x$ from $1$ to $1.5$ . A GDC table or spreadsheet does this quickly.
Round only at the very end, to three decimal places.

y (1.1) = 4 + 0.1 (1 + ln 2) = 4.16931 \dots y (1.2) = 4.35350 \dots, y (1.3) = 4.55235 \dots y (1.4) = 4.76564 \dots, y (1.5) = 4.99319 \dots y (1.5) \approx 4.993

Mentor tip

Each step uses the gradient at the start of the step. Count five steps, and keep full accuracy between them.

Question 8

7 marks

Loan repayments with the finance solver

Number and algebra

(a)

3 marks

Enter the loan into the finance solver: $60$ monthly periods, $8.4%$ nominal, compounded $12$ times a year.
Loan in, repayments out, so $P V$ and $P M T$ have opposite signs.

N = 60, I % = 8.4, P V = 30000, F V = 0, P / Y = C / Y = 12 P M T = - 614.05 (GDC) Monthly payment = $614.05

Mentor tip

Give the amount as a positive number of dollars to 2 d.p. The minus sign is only the solver's cash-flow convention.

(b)

4 marks

Find the balance still owed after $24$ payments: run the solver with $N = 24$ and solve for $F V$ .
Subtract the 5000 dollars to get the new balance.
Use that as the new $P V$ over $N = 36$ months with $F V = 0$ and solve for $P M T$ .

N = 24, I % = 8.4, P V = 30000, P M T = - 614.05 \dots ⟹ F V = - 19 480.56 (GDC) 19 480.56 - 5000 = 14 480.56 N = 36, P V = 14 480.56, F V = 0 ⟹ P M T = - 456.44 (GDC) New monthly payment = $456.44

Mentor tip

The interest rate stays the same after the lump sum. Only the balance and the number of months change.

Question 9

8 marks

Adjacency matrix and counting walks

Number and algebra

(a)

3 marks

(i) Order the rows and columns $A, B, C, D$ . Put a $1$ in row $i$ , column $j$ when there is an arrow from $i$ to $j$ .

The arrows are $A \to B$ , $A \to D$ , $B \to C$ , $D \to C$ and $C \to A$ .

(ii) Enter $M$ on the GDC and raise it to the power $4$ .

M = 0010100001011000 M^{4} = 0020200002022000 (GDC)

Mentor tip

The diagonal is one-way, from $C$ to $A$ . Check each arrow's direction before filling in the matrix.

(b)

1 mark

Read row $A$ of $M^{4}$ : the non-zero entries mark the reachable end vertices.

B and D

(c)

2 marks

Raise $M$ to the powers $7$ and $10$ on the GDC. Notice each result is $M$ times a power of $2$ .

(i) M^{7} = 0040400004044000 (ii) M^{10} = 0080800008088000

(d)

2 marks

The powers $1, 4, 7, 10$ are $3 n - 2$ for $n = 1, 2, 3, 4$ , and the matrices are $M, 2 M, 4 M, 8 M$ .
So $M^{3 n - 2} = 2^{n - 1} M$ . The $D$ to $C$ entry of $M$ is $1$ .

M^{3 n - 2} = 2^{n - 1} M walks from D to C = 2^{n - 1}

Mentor tip

Why it works: $M^{4} = 2 M$ , so every extra three steps doubles each entry. Note $M^{3}$ itself is not $2 I$ : only $A$ and $C$ have two 3-step loops back to themselves.

Question 10

8 marks

Correlation coefficient and a test for correlation

Statistics and probability

(a)

2 marks

Enter hours as $x$ and scores as $y$ in two lists and run linear regression on the GDC.

r = 0.692116 \dots \approx 0.692 (GDC)

Mentor tip

Give $r$ to 3 s.f. Two figures, $0.69$ , is not accurate enough.

(b)

6 marks

(i) The test is about the population correlation $ρ$ . The claim is a positive correlation, so the test is one-tailed.

(ii) Run the linear regression $t$ -test on the GDC with the alternative $ρ > 0$ .

(iii) Compare the $p$ -value with $0.05$ and conclude.

(i) H_{0} : ρ = 0, H_{1} : ρ > 0 (ii) p = 0.0285759 \dots \approx 0.0286 (GDC) (iii) 0.0286 < 0.05 ⟹ reject H_{0} There is evidence of a positive correlation between study time and score.

Mentor tip

Set the GDC to the one-sided option. The two-sided setting doubles the $p$ -value to $0.0572$ and flips the conclusion.

Question 11

6 marks

Kruskal's algorithm for a minimum spanning tree

Geometry and trigonometry

(a)

4 marks

Seven vertices need six edges. List the edges from cheapest upwards.
Add each edge in turn unless it would close a cycle.
The two cost-9 edges can go in either order. Both cost-10 edges get skipped because they would make cycles.

A D (4), C E (5), D F (7), B E (9), A B (9) B C (10) rejected (cycle B - C - E), E F (10) rejected (cycle E - F - D - A - B - E) E G (11) added: six edges, stop

Mentor tip

Write the edges in the order you consider them, including the rejected ones. The order is the evidence you used Kruskal's.

(b)

1 mark

Draw all seven vertices and only the six chosen edges: $A D$ , $C E$ , $D F$ , $B E$ , $A B$ and $E G$ .
The result is one connected tree: $F - D - A - B - E$ , with $C$ and $G$ each hanging off $E$ .

edges: A D, A B, B E, C E, D F, E G

(c)

1 mark

Add the six edge weights and give the units.

4 + 5 + 7 + 9 + 9 + 11 = 45 $45 million

Question 12

9 marks

Discrete random variable: k, mean and repeats

Statistics and probability

(a)

3 marks

The probabilities must add to $1$ .
Work out each probability in terms of $k$ for $x = 1, 2, 3$ and add.

k (1) (15) + k (2) (12) + k (3) (7) = 1 15 k + 24 k + 21 k = 60 k = 1 k = \frac{1}{60}

Mentor tip

On a show-that, write every probability out in full before collecting terms.

(b)

3 marks

(i) Substitute $x = 2$ with $k = \frac{1}{60}$ .

(ii) Use $E (X) = \sum x P (X = x)$ .

(i) P (X = 2) = \frac{24}{60} = 0.4 (ii) E (X) = 1 \cdot \frac{15}{60} + 2 \cdot \frac{24}{60} + 3 \cdot \frac{21}{60} = \frac{126}{60} = 2.1

(c)

3 marks

The two values are equal if both are $1$ , both are $2$ or both are $3$ .
Independence means each case is a probability squared. Add the three cases.

(\frac{15}{60})^{2} + (\frac{24}{60})^{2} + (\frac{21}{60})^{2} = \frac{225 + 576 + 441}{3600} = \frac{1242}{3600} = \frac{69}{200} = 0.345

Mentor tip

Do not forget any of the three matching cases. A quick sense check: the answer must be below the largest single probability, $0.4$ .

Question 13

8 marks

Kinematics with velocity, distance and acceleration

Calculus

(a)

2 marks

At rest means $v = 0$ . Since $e^{- t}$ is never zero, solve $t^{2} (3 - t) = 0$ .

(3 t^{2} - t^{3}) e^{- t} = 0 ⟹ t^{2} (3 - t) = 0 t = 3 s

(b)

2 marks

Distance is the integral of $∣ v ∣$ from $0$ to $3$ . Here $v \geq 0$ on the whole interval, so a plain integral works too.

\int_{0}^{3} (3 t^{2} - t^{3}) e^{- t} d t = 1.34425 \dots \approx 1.34 m (GDC)

Mentor tip

Always write the integral with its limits before giving the GDC value.

(c)

2 marks

Differentiate $v$ with the product rule: $u = 3 t^{2} - t^{3}$ and $w = e^{- t}$ .

a = (6 t - 3 t^{2}) e^{- t} - (3 t^{2} - t^{3}) e^{- t} a = (t^{3} - 6 t^{2} + 6 t) e^{- t}

Mentor tip

The derivative of $e^{- t}$ is $- e^{- t}$ . The minus sign is the usual slip.

(d)

2 marks

Graph $a (t)$ from part (c) on the GDC for $t \geq 0$ and find its local maximum.

t = 0.41577 \dots \approx 0.416 s (GDC)

Another way. Set $a^{'} (t) = (- t^{3} + 9 t^{2} - 18 t + 6) e^{- t} = 0$ and take the smallest positive root.

Mentor tip

Maximum acceleration is where the velocity graph is steepest going up, not where the velocity itself is largest.

Question 14

8 marks

Complex conjugates, modulus and argument

Number and algebra

(a)

1 mark

The conjugate keeps the real part and flips the sign of the imaginary part.

z^{*} = a - b i

(b)

3 marks

Expand the brackets and use $i^{2} = - 1$ .
The imaginary terms cancel, leaving $a^{2} + b^{2}$ .
Since $a$ and $b$ are real, $a^{2} + b^{2}$ is real.

z z^{*} = (a + b i) (a - b i) = a^{2} - ab i + ab i - b^{2} i^{2} = a^{2} + b^{2} \in R

Mentor tip

Finish with a sentence explaining why the result is real. The reason is part of the answer.

(c)

2 marks

From (b), $a^{2} + b^{2} = 5$ . With $a$ and $b$ positive integers, the only options are $1$ and $2$ in some order.

a^{2} + b^{2} = 5 ⟹ (a, b) = (1, 2) or (2, 1) z_{1} = 1 + 2 i, z_{2} = 2 + i

Mentor tip

Give just these two. Listing the conjugates as well adds wrong answers.

(d)

2 marks

Both numbers are in the first quadrant, so each argument is $arctan \frac{b}{a}$ .
Alternatively, the arguments add when you multiply: $z_{1} z_{2} = 5 i$ , which points straight up.

ar g z_{1} + ar g z_{2} = arctan 2 + arctan 0.5 = ar g (z_{1} z_{2}) = ar g (5 i) = \frac{π}{2} (\approx 1.57)

Mentor tip

Your GDC can check this: add the two arctangents in radian mode and you get $1.5707 \dots$ .

Question 15

6 marks

Route inspection on a weighted graph

Geometry and trigonometry

(a)

1 mark

Count the edges meeting at $T$ : $T S$ and $T U$ .

de g (T) = 2

(b)

1 mark

A closed route using every edge once needs every vertex to have even degree. Here $P$ , $Q$ , $S$ and $U$ have odd degree.

de g P = 3, de g Q = 3, de g S = 5, de g U = 3 Not all degrees are even, so the graph is not Eulerian.

(c)

4 marks

(i) Pair up the four odd vertices $P, Q, S, U$ in all three ways and find the shortest route for each pair.

Choose the pairing with the smallest total and repeat those edges.

(ii) Add the repeated distance to the total of all edge weights.

QS + P U = 18 + 16 = 34 Q U + P S = 37 + 23 = 60, QP + S U = 21 + 19 = 40 (i) repeat QS and P U all edges = 21 + 23 + 18 + 19 + 14 + 20 + 16 + 19 + 13 = 163 (ii) 163 + 34 = 197 km

Mentor tip

For $Q U$ there is no direct edge, so use the shortest path ( $Q$ to $S$ to $U$ or $Q$ to $P$ to $U$ , both $37$ ).

Question 16

9 marks

Trig graphs, an identity and a volume of revolution

Functions
Calculus

(a)

2 marks

(i) $g$ : amplitude $0.5$ and period $π$ . Maxima of $0.5$ at $x = 0, π, 2 π$ ; minima of $- 0.5$ at $x = \frac{π}{2}, \frac{3 π}{2}$ ; it crosses the $x$ -axis at $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ .

(ii) $h$ : the same wave shape, but never negative. Maxima of $1$ at $x = 0, π, 2 π$ ; minima of $0$ (touching the $x$ -axis) at $x = \frac{π}{2}, \frac{3 π}{2}$ ; it equals $0.5$ at the odd multiples of $\frac{π}{4}$ .

g : - 0.5 \leq y \leq 0.5, period π h : 0 \leq y \leq 1, period π

Mentor tip

Plot both on the GDC first and copy the turning points onto the grid carefully, one small square at a time.

(b)

1 mark

$h$ has the same shape as $g$ but sits $0.5$ higher: compare the maxima, $1$ against $0.5$ .

h (x) = g (x) + 0.5

Mentor tip

This is the double angle identity $cos^{2} x = \frac{1}{2} + \frac{1}{2} cos 2 x$ in disguise.

(c)

6 marks

(i) Use $V = π \int y^{2} d x$ with $y = cos x$ between $0$ and $\frac{π}{2}$ .

(ii) Replace $cos^{2} x$ with $0.5 + 0.5 cos 2 x$ from part (b).

Integrate: $0.5 cos 2 x$ integrates to $0.25 sin 2 x$ .

Substitute the limits; both sine terms are zero.

(i) V = π \int_{0}^{\frac{π}{2}} cos^{2} x d x (ii) V = π \int_{0}^{\frac{π}{2}} (0.5 + 0.5 cos 2 x) d x = π [0.5 x + 0.25 sin 2 x]_{0}^{\frac{π}{2}} = π (\frac{π}{4} + 0.25 sin π) - π (0 + 0) = \frac{π ^{2}}{4}

Mentor tip

Keep the $π$ in front of the integral all the way through. Dropping it is the most common reason the answer comes out as $\frac{π}{4}$ .

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra34 marks · 31%
Q4, Q5, Q8, Q9, Q14
Statistics and probability26 marks · 24%
Q3, Q10, Q12
Geometry and trigonometry24 marks · 22%
Q2, Q6, Q11, Q15
Calculus18 marks · 16%
Q7, Q13, Q16
Functions8 marks · 7%
Q1, Q16

Mentor insights

Traps to watch for

01
Arrow direction on the diagonal · Q9
The diagonal only goes from C to A. Treating it as two-way puts an extra 1 in the matrix and every later power goes wrong.
02
Two-tailed p-value for a one-sided claim · Q10
The claim is a positive correlation, so use the one-sided test. The two-sided p-value is about 0.057, which would wrongly fail to reject at 5%.
03
Wrong denominator in the conditional · Q3
Given over 50 means dividing by the over-50 column total of 53, not by 180.
04
Rushing the balance after two years · Q8
Find the outstanding balance after 24 payments with the finance solver first, then take off the 5000 before working out the new payment.
05
Missing a pairing of odd vertices · Q15
Check all three ways to pair P, Q, S and U. Use shortest paths where there is no direct edge, then pick the smallest total.

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Worked solutions

IB Maths AI HL · Paper 1

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2026
Marks: 110
Duration: 2 hours
Calculator: Allowed
Questions: 16

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–16 · 110 marks

Question 1

5 marks

Exponential population model

Functions

(a)

1 mark

At the start of 2024, $t = 0$ and $b^{0} = 1$ , so $a$ is just the starting population.

f (0) = a \times b^{0} = a a = 104000

Mentor tip

The value of $a$ in $a \times b^{t}$ is always the value at $t = 0$ .

(b)

2 marks

One year later $t = 1$ . Set $f (1)$ equal to the 2025 figure and divide.

104000 \times b = 105144 b = \frac{105 144}{104 000} = 1.011

Mentor tip

The word exact matters here: give $1.011$ in full, not a rounded value.

(c)

2 marks

The start of 2034 is ten years on, so use $t = 10$ .
Round to the nearest whole person, as asked.

f (10) = 104000 \times 1.01 1^{10} = 116 023.2 \dots \approx 116023

Mentor tip

Count the years carefully: 2024 is $t = 0$ , so 2034 is $t = 10$ , not $t = 11$ .

Question 2

5 marks

Midpoint and distance in 3D

Geometry and trigonometry

(a)

2 marks

Average each coordinate of $A$ and $B$ .

F = (\frac{0 + 40}{2}, \frac{60 + 0}{2}, \frac{240 + 240}{2}) F = (20, 30, 240)

Mentor tip

Write the answer as a bracketed coordinate triple, not three loose numbers.

(b)

3 marks

Use the 3D distance formula between $S (40, 0, 160)$ and $F (20, 30, 240)$ .

S F = (40 - 20)^{2} + (0 - 30)^{2} + (160 - 240)^{2} = 400 + 900 + 6400 = 7700 \approx 87.7 cm

Another way. Find $F B = 2 0^{2} + 3 0^{2}$ across the level of the ceiling thread, then combine it with the vertical drop $S B = 80$ using Pythagoras.

Mentor tip

The distance formula is in the formula booklet. The squares remove any sign worries.

Question 3

9 marks

Two-way table and a chi-squared test

Statistics and probability

(a)

1 mark

Read the single cell for under 30 and cold, out of all $180$ people.

\frac{62}{180} = \frac{31}{90} \approx 0.344

(b)

2 marks

The condition shrinks the group to the over-50 column only: $19 + 34 = 53$ people.
Of those, $19$ prefer hot.

\frac{19}{53} \approx 0.358

Mentor tip

For a conditional probability the denominator is the column total, not $180$ .

(c)

1 mark

The null hypothesis for a test of independence always says the two variables are independent.

H_{0} : age and preferred type of drink are independent

Mentor tip

Use the word independent. Saying the variables are not related or not correlated is too loose.

(d)

3 marks

Enter the $2 \times 3$ table of observed counts as a matrix and run the chi-squared test on the GDC.
Read off both the statistic and the $p$ -value. There are $(2 - 1) (3 - 1) = 2$ degrees of freedom.

(i) χ_{calc}^{2} = 3.65997 \dots \approx 3.66 (GDC) (ii) p = 0.160415 \dots \approx 0.160 (GDC)

Mentor tip

Enter only the six observed counts, not the row or column totals.

(e)

2 marks

Compare the statistic with the critical value (or the $p$ -value with $0.01$ ).
The statistic is smaller, so there is not enough evidence to reject $H_{0}$ .

3.66 < 9.210 (or 0.160 > 0.01) Do not reject H_{0} : age and drink preference are independent.

Mentor tip

Always write the comparison first, then the conclusion. A conclusion with no reason is only half an answer.

Question 4

4 marks

Linear model from two bills

Number and algebra

4 marks

The bill is linear: bill $= c + u \times$ (units used).
Write one equation for each month and solve them simultaneously (GDC or subtraction).
(i) and (ii): subtracting the equations removes $c$ and gives $u$ ; substitute back for $c$ .

c + 3000 u = 117.90 c + 4800 u = 179.10 1800 u = 61.20 ⟹ u = 0.034 c = 117.90 - 3000 (0.034) = 15.90 (i) c = $15.90 (ii) u = $0.034

Mentor tip

Check both bills with your values: $15.90 + 4800 \times 0.034 = 179.10$ .

Question 5

7 marks

Arithmetic sequence: a term and a sum

Number and algebra

(a)

2 marks

Use $u_{n} = u_{1} + (n - 1) d$ with $u_{1} = 18$ , $d = 7$ and $n = 42$ .

u_{42} = 18 + 41 \times 7 = 305

Mentor tip

It is $41$ steps of $7$ to reach the 42nd term, not $42$ .

(b)

3 marks

Put the values into $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ and set it equal to $9475$ .
Expand and move everything to one side.

\frac{n}{2} (36 + 7 (n - 1)) = 9475 \frac{n}{2} (7 n + 29) = 9475 7 n^{2} + 29 n - 18950 = 0

Mentor tip

Halving every term, $3.5 n^{2} + 14.5 n - 9475 = 0$ , is just as good.

(c)

2 marks

Solve the quadratic with the GDC polynomial solver.
Keep only the positive whole-number root, since $n$ counts terms.

n = 50 or n = - 54.14 \dots (GDC) n = 50

Mentor tip

Reject the negative root in writing and give one final answer.

Question 6

7 marks

Closest point on a line to a point

Geometry and trigonometry

7 marks

The closest point is where a perpendicular from $C$ meets the road.
Find the gradient and equation of the road through $A$ and $B$ .
The perpendicular through $C$ has the negative reciprocal gradient.
Solve the two line equations simultaneously to get $D$ .

m_{A B} = \frac{2 - 10}{5 - 1} = - 2 road: y = - 2 x + 12 m_{C D} = \frac{1}{2} : y - 5 = \frac{1}{2} (x - 4) ⟹ y = 0.5 x + 3 - 2 x + 12 = 0.5 x + 3 ⟹ x = 3.6 D = (3.6, 4.8)

Another way. Write the distance from $C$ to a general point $(x, - 2 x + 12)$ on the road and use the GDC to find the $x$ that minimises it.

Mentor tip

Check $D$ lies on the road: $- 2 (3.6) + 12 = 4.8$ .

Question 7

4 marks

Euler's method for a differential equation

Calculus

4 marks

Use $y_{n + 1} = y_{n} + 0.1 (x_{n} + ln (1 + x_{n}))$ and $x_{n + 1} = x_{n} + 0.1$ , starting at $(1, 4)$ .
Five steps take $x$ from $1$ to $1.5$ . A GDC table or spreadsheet does this quickly.
Round only at the very end, to three decimal places.

y (1.1) = 4 + 0.1 (1 + ln 2) = 4.16931 \dots y (1.2) = 4.35350 \dots, y (1.3) = 4.55235 \dots y (1.4) = 4.76564 \dots, y (1.5) = 4.99319 \dots y (1.5) \approx 4.993

Mentor tip

Each step uses the gradient at the start of the step. Count five steps, and keep full accuracy between them.

Question 8

7 marks

Loan repayments with the finance solver

Number and algebra

(a)

3 marks

Enter the loan into the finance solver: $60$ monthly periods, $8.4%$ nominal, compounded $12$ times a year.
Loan in, repayments out, so $P V$ and $P M T$ have opposite signs.

N = 60, I % = 8.4, P V = 30000, F V = 0, P / Y = C / Y = 12 P M T = - 614.05 (GDC) Monthly payment = $614.05

Mentor tip

Give the amount as a positive number of dollars to 2 d.p. The minus sign is only the solver's cash-flow convention.

(b)

4 marks

Find the balance still owed after $24$ payments: run the solver with $N = 24$ and solve for $F V$ .
Subtract the 5000 dollars to get the new balance.
Use that as the new $P V$ over $N = 36$ months with $F V = 0$ and solve for $P M T$ .

N = 24, I % = 8.4, P V = 30000, P M T = - 614.05 \dots ⟹ F V = - 19 480.56 (GDC) 19 480.56 - 5000 = 14 480.56 N = 36, P V = 14 480.56, F V = 0 ⟹ P M T = - 456.44 (GDC) New monthly payment = $456.44

Mentor tip

The interest rate stays the same after the lump sum. Only the balance and the number of months change.

Question 9

8 marks

Adjacency matrix and counting walks

Number and algebra

(a)

3 marks

(i) Order the rows and columns $A, B, C, D$ . Put a $1$ in row $i$ , column $j$ when there is an arrow from $i$ to $j$ .

The arrows are $A \to B$ , $A \to D$ , $B \to C$ , $D \to C$ and $C \to A$ .

(ii) Enter $M$ on the GDC and raise it to the power $4$ .

M = 0010100001011000 M^{4} = 0020200002022000 (GDC)

Mentor tip

The diagonal is one-way, from $C$ to $A$ . Check each arrow's direction before filling in the matrix.

(b)

1 mark

Read row $A$ of $M^{4}$ : the non-zero entries mark the reachable end vertices.

B and D

(c)

2 marks

Raise $M$ to the powers $7$ and $10$ on the GDC. Notice each result is $M$ times a power of $2$ .

(i) M^{7} = 0040400004044000 (ii) M^{10} = 0080800008088000

(d)

2 marks

The powers $1, 4, 7, 10$ are $3 n - 2$ for $n = 1, 2, 3, 4$ , and the matrices are $M, 2 M, 4 M, 8 M$ .
So $M^{3 n - 2} = 2^{n - 1} M$ . The $D$ to $C$ entry of $M$ is $1$ .

M^{3 n - 2} = 2^{n - 1} M walks from D to C = 2^{n - 1}

Mentor tip

Why it works: $M^{4} = 2 M$ , so every extra three steps doubles each entry. Note $M^{3}$ itself is not $2 I$ : only $A$ and $C$ have two 3-step loops back to themselves.

Question 10

8 marks

Correlation coefficient and a test for correlation

Statistics and probability

(a)

2 marks

Enter hours as $x$ and scores as $y$ in two lists and run linear regression on the GDC.

r = 0.692116 \dots \approx 0.692 (GDC)

Mentor tip

Give $r$ to 3 s.f. Two figures, $0.69$ , is not accurate enough.

(b)

6 marks

(i) The test is about the population correlation $ρ$ . The claim is a positive correlation, so the test is one-tailed.

(ii) Run the linear regression $t$ -test on the GDC with the alternative $ρ > 0$ .

(iii) Compare the $p$ -value with $0.05$ and conclude.

(i) H_{0} : ρ = 0, H_{1} : ρ > 0 (ii) p = 0.0285759 \dots \approx 0.0286 (GDC) (iii) 0.0286 < 0.05 ⟹ reject H_{0} There is evidence of a positive correlation between study time and score.

Mentor tip

Set the GDC to the one-sided option. The two-sided setting doubles the $p$ -value to $0.0572$ and flips the conclusion.

Question 11

6 marks

Kruskal's algorithm for a minimum spanning tree

Geometry and trigonometry

(a)

4 marks

Seven vertices need six edges. List the edges from cheapest upwards.
Add each edge in turn unless it would close a cycle.
The two cost-9 edges can go in either order. Both cost-10 edges get skipped because they would make cycles.

A D (4), C E (5), D F (7), B E (9), A B (9) B C (10) rejected (cycle B - C - E), E F (10) rejected (cycle E - F - D - A - B - E) E G (11) added: six edges, stop

Mentor tip

Write the edges in the order you consider them, including the rejected ones. The order is the evidence you used Kruskal's.

(b)

1 mark

Draw all seven vertices and only the six chosen edges: $A D$ , $C E$ , $D F$ , $B E$ , $A B$ and $E G$ .
The result is one connected tree: $F - D - A - B - E$ , with $C$ and $G$ each hanging off $E$ .

edges: A D, A B, B E, C E, D F, E G

(c)

1 mark

Add the six edge weights and give the units.

4 + 5 + 7 + 9 + 9 + 11 = 45 $45 million

Question 12

9 marks

Discrete random variable: k, mean and repeats

Statistics and probability

(a)

3 marks

The probabilities must add to $1$ .
Work out each probability in terms of $k$ for $x = 1, 2, 3$ and add.

k (1) (15) + k (2) (12) + k (3) (7) = 1 15 k + 24 k + 21 k = 60 k = 1 k = \frac{1}{60}

Mentor tip

On a show-that, write every probability out in full before collecting terms.

(b)

3 marks

(i) Substitute $x = 2$ with $k = \frac{1}{60}$ .

(ii) Use $E (X) = \sum x P (X = x)$ .

(i) P (X = 2) = \frac{24}{60} = 0.4 (ii) E (X) = 1 \cdot \frac{15}{60} + 2 \cdot \frac{24}{60} + 3 \cdot \frac{21}{60} = \frac{126}{60} = 2.1

(c)

3 marks

The two values are equal if both are $1$ , both are $2$ or both are $3$ .
Independence means each case is a probability squared. Add the three cases.

(\frac{15}{60})^{2} + (\frac{24}{60})^{2} + (\frac{21}{60})^{2} = \frac{225 + 576 + 441}{3600} = \frac{1242}{3600} = \frac{69}{200} = 0.345

Mentor tip

Do not forget any of the three matching cases. A quick sense check: the answer must be below the largest single probability, $0.4$ .

Question 13

8 marks

Kinematics with velocity, distance and acceleration

Calculus

(a)

2 marks

At rest means $v = 0$ . Since $e^{- t}$ is never zero, solve $t^{2} (3 - t) = 0$ .

(3 t^{2} - t^{3}) e^{- t} = 0 ⟹ t^{2} (3 - t) = 0 t = 3 s

(b)

2 marks

Distance is the integral of $∣ v ∣$ from $0$ to $3$ . Here $v \geq 0$ on the whole interval, so a plain integral works too.

\int_{0}^{3} (3 t^{2} - t^{3}) e^{- t} d t = 1.34425 \dots \approx 1.34 m (GDC)

Mentor tip

Always write the integral with its limits before giving the GDC value.

(c)

2 marks

Differentiate $v$ with the product rule: $u = 3 t^{2} - t^{3}$ and $w = e^{- t}$ .

a = (6 t - 3 t^{2}) e^{- t} - (3 t^{2} - t^{3}) e^{- t} a = (t^{3} - 6 t^{2} + 6 t) e^{- t}

Mentor tip

The derivative of $e^{- t}$ is $- e^{- t}$ . The minus sign is the usual slip.

(d)

2 marks

Graph $a (t)$ from part (c) on the GDC for $t \geq 0$ and find its local maximum.

t = 0.41577 \dots \approx 0.416 s (GDC)

Another way. Set $a^{'} (t) = (- t^{3} + 9 t^{2} - 18 t + 6) e^{- t} = 0$ and take the smallest positive root.

Mentor tip

Maximum acceleration is where the velocity graph is steepest going up, not where the velocity itself is largest.

Question 14

8 marks

Complex conjugates, modulus and argument

Number and algebra

(a)

1 mark

The conjugate keeps the real part and flips the sign of the imaginary part.

z^{*} = a - b i

(b)

3 marks

Expand the brackets and use $i^{2} = - 1$ .
The imaginary terms cancel, leaving $a^{2} + b^{2}$ .
Since $a$ and $b$ are real, $a^{2} + b^{2}$ is real.

z z^{*} = (a + b i) (a - b i) = a^{2} - ab i + ab i - b^{2} i^{2} = a^{2} + b^{2} \in R

Mentor tip

Finish with a sentence explaining why the result is real. The reason is part of the answer.

(c)

2 marks

From (b), $a^{2} + b^{2} = 5$ . With $a$ and $b$ positive integers, the only options are $1$ and $2$ in some order.

a^{2} + b^{2} = 5 ⟹ (a, b) = (1, 2) or (2, 1) z_{1} = 1 + 2 i, z_{2} = 2 + i

Mentor tip

Give just these two. Listing the conjugates as well adds wrong answers.

(d)

2 marks

Both numbers are in the first quadrant, so each argument is $arctan \frac{b}{a}$ .
Alternatively, the arguments add when you multiply: $z_{1} z_{2} = 5 i$ , which points straight up.

ar g z_{1} + ar g z_{2} = arctan 2 + arctan 0.5 = ar g (z_{1} z_{2}) = ar g (5 i) = \frac{π}{2} (\approx 1.57)

Mentor tip

Your GDC can check this: add the two arctangents in radian mode and you get $1.5707 \dots$ .

Question 15

6 marks

Route inspection on a weighted graph

Geometry and trigonometry

(a)

1 mark

Count the edges meeting at $T$ : $T S$ and $T U$ .

de g (T) = 2

(b)

1 mark

A closed route using every edge once needs every vertex to have even degree. Here $P$ , $Q$ , $S$ and $U$ have odd degree.

de g P = 3, de g Q = 3, de g S = 5, de g U = 3 Not all degrees are even, so the graph is not Eulerian.

(c)

4 marks

(i) Pair up the four odd vertices $P, Q, S, U$ in all three ways and find the shortest route for each pair.

Choose the pairing with the smallest total and repeat those edges.

(ii) Add the repeated distance to the total of all edge weights.

QS + P U = 18 + 16 = 34 Q U + P S = 37 + 23 = 60, QP + S U = 21 + 19 = 40 (i) repeat QS and P U all edges = 21 + 23 + 18 + 19 + 14 + 20 + 16 + 19 + 13 = 163 (ii) 163 + 34 = 197 km

Mentor tip

For $Q U$ there is no direct edge, so use the shortest path ( $Q$ to $S$ to $U$ or $Q$ to $P$ to $U$ , both $37$ ).

Question 16

9 marks

Trig graphs, an identity and a volume of revolution

Functions
Calculus

(a)

2 marks

g : - 0.5 \leq y \leq 0.5, period π h : 0 \leq y \leq 1, period π

Mentor tip

Plot both on the GDC first and copy the turning points onto the grid carefully, one small square at a time.

(b)

1 mark

$h$ has the same shape as $g$ but sits $0.5$ higher: compare the maxima, $1$ against $0.5$ .

h (x) = g (x) + 0.5

Mentor tip

This is the double angle identity $cos^{2} x = \frac{1}{2} + \frac{1}{2} cos 2 x$ in disguise.

(c)

6 marks

(i) Use $V = π \int y^{2} d x$ with $y = cos x$ between $0$ and $\frac{π}{2}$ .

(ii) Replace $cos^{2} x$ with $0.5 + 0.5 cos 2 x$ from part (b).

Integrate: $0.5 cos 2 x$ integrates to $0.25 sin 2 x$ .

Substitute the limits; both sine terms are zero.

(i) V = π \int_{0}^{\frac{π}{2}} cos^{2} x d x (ii) V = π \int_{0}^{\frac{π}{2}} (0.5 + 0.5 cos 2 x) d x = π [0.5 x + 0.25 sin 2 x]_{0}^{\frac{π}{2}} = π (\frac{π}{4} + 0.25 sin π) - π (0 + 0) = \frac{π ^{2}}{4}

Mentor tip

Keep the $π$ in front of the integral all the way through. Dropping it is the most common reason the answer comes out as $\frac{π}{4}$ .

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra34 marks · 31%
Q4, Q5, Q8, Q9, Q14
Statistics and probability26 marks · 24%
Q3, Q10, Q12
Geometry and trigonometry24 marks · 22%
Q2, Q6, Q11, Q15
Calculus18 marks · 16%
Q7, Q13, Q16
Functions8 marks · 7%
Q1, Q16

Mentor insights

Traps to watch for

01
Arrow direction on the diagonal · Q9
The diagonal only goes from C to A. Treating it as two-way puts an extra 1 in the matrix and every later power goes wrong.
02
Two-tailed p-value for a one-sided claim · Q10
The claim is a positive correlation, so use the one-sided test. The two-sided p-value is about 0.057, which would wrongly fail to reject at 5%.
03
Wrong denominator in the conditional · Q3
Given over 50 means dividing by the over-50 column total of 53, not by 180.
04
Rushing the balance after two years · Q8
Find the outstanding balance after 24 payments with the finance solver first, then take off the 5000 before working out the new payment.
05
Missing a pairing of odd vertices · Q15
Check all three ways to pair P, Q, S and U. Use shortest paths where there is no direct edge, then pick the smallest total.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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